c++ - 从 std::shared_ptr 初始化 std::shared_ptr

Question

我在尝试从 std::shared_ptr<void> 中“复制”时遇到了它模板化 std::shared_ptr<T> .它看起来像这样:

template < >
class TDynamic< void >
{
protected:
    std::shared_ptr< void > m_pointer;
public:
    // Here i got constructors, functions etc
    template < typename U >
    TDynamic< U > SwitchType()
    {
        TDynamic< U > returnValue;
        returnValue.m_pointer = std::shared_ptr< U >(m_pointer); // error here
    }
};

错误是这样说的:

c:\...path_here...\tdynamic.hpp(426): error C2440: 'type cast': cannot convert from 'const std::shared_ptr< void >' to 'std::shared_ptr< T >'
1>          with
1>          [
1>              T=grim::Actor
1>          ]
1>  c:\...path_here...\tdynamic.hpp(426): note: No constructor could take the source type, or constructor overload resolution was ambiguous

编译器和 IDE:Visual Studio 2015 社区

对于这种情况，这对我来说是必要的。假设我有类“Actor”和派生类“APlayer”

class Actor
{
public:
};
class APlayer
    : public Actor
{
public:
};

int main()
{ // some code here
    TDynamic< APlayer > test1(new APlayer);
    TDynamic< void > test2(test1);
    TDynamic< Actor > test3(test2); // error here
}

我可以将整个代码粘贴到此处，但大约有 550 行，其中大部分都无关紧要。我只需要复制那个 std::shared_ptr .

Answer 1

您可以使用 static_pointer_cast :

returnValue.m_pointer = static_pointer_cast<U>(m_pointer);

由您来保证此转换是合法的(它被定义为与 static_cast<U*>(m_pointer.get()) 一样合法)。